Determine the force in members BE, BJ, DJ, and IJ. (ADJ and HDB are not continues, each consist of two elements.)

The first step is to find the support reaction forces. Vertical reaction force at F can be found by taking the moment about L.

     ΣM_L= 0
            = - 6F_y + 5G_y + 4H_y + 3I_y + 2J_y
            = - 6F_y + 5(2) + 4(3) + 3(5) + 2(2)
            = - 6F_y + 41 = 0

     F_y = 6.833 kN

Summation of forces in the y direction should be zero.

     ΣF= 0
     F_y + L_y - (G_y + H_y + I_y + J_y) = 0
     6.833 + L_y - (2 + 3 + 5 + 2) = 0
     L_y = 5.167 kN

Another method to find L_y is to take the moment about F. It is a good way to check to see if the previous value for L_y was found correctly.

     ΣM_F = 0
             = 6L_y - ( G_y + 2H_y + 3I_y + 4J_y )
             = 6L_y - ( 2 + 2(3) + 3(5) + 4(2) )
             = 6L_y - 31 = 0

     L_y = 5.167 kN

To find the value of the forces it is easier to use method of sections. Since the problem asks for four member forces, BE, BJ, DJ, and IJ, the problem needs to be solved taking two cuts. The first cut , shown at the left, will be used to solve for member force in member BE. Then another cut will be used to find the other three member forces.

Moments about J can be taken to find the value of the F_BE (notation EJ represents the length of the member EJ).

      ΣM_J = 0
              = 2L + F_BE(EJ)

JBE is a right triangle and BE is perpendicular to EJ. The length of KE and JK is 1. Based on Pythagorean theorem, EJ is the square root of 2. (EJ = 1.414).

     2L + F_BE(1.414) = 0
     2(5.167) + F_BE(1.414) = 0
     F_EB = -7.308 kN

Now that F_BE is known, another cut can be made to find F_JB, F_JD, and F_IJ. To find F_JD and F_JB, moment about L can be used with the sum of the vertical forces (notation JL mean length of JL)

      ΣM_L = 0
               = JL (2kN) - F_JD(BD) - F_JB(JL)
               = 2(2) - F_JD(1.414) - F_JB(2)

     ΣF_y = 0
            = F_JD + F_JB + 5.167 - 2
                       -7.308(sin45)

F_JD and F_JB can be found by solving the above system of equations to give

      F_JD = -0.8327(10^-3) kN    (very small)

      F_JB = 2.000 kN

F_JD is found to be 0.8 newton which is negligible and is because of the rounding errors.

Summation of forces in x direction can be written in order to find the value of F_IJ

      ΣF_x = 0
             = F_EB(cos45) + F_JD(cos45) + F_IJ
             = -7.308 (0.707) + 0 (0.707) + F_IJ

      F_IJ = 5.167 kN

F_IJ can also be found by taking the moment about joint B

     ΣM_B = 0
               = L(JL) - F_JD(BD) - F_IJ(BJ)
               = 5.167(2) + 0 (1.414) - F_IJ(2)

     F_IJ = 5.167 kN